Question

If -3 is a root of the quadratic equation 2x² +px-15=0 and the quadratic equation p(x²+x) +k=0 has equal routs, find the value of K.​

Answer 1

Given: -3 is the root of the quadratic equation $2x^{2} +px-15=0$.

And the equation  $p(x^{2} +x) +k=0$  has equal roots.

To find: Value of k

Solution:

Since -3 is the root of the equation, it does satisfy the equation.

So, substitute -3 in the equation ,$2x^{2} +px-15=0$, in order to get the value of p.

⇒ 2(9) - 3p - 15 = 0

⇒ 18 - 3p - 15 = 0

⇒ 3p = 3

⇒ p = 1

The value of p is 1

If the roots of a quadratic equation are equal then $b^{2} = 4ac$

By applying this condition on $p(x^{2} +x) +k=0$, we get

⇒ $p^{2}x^{2} = 4pk$

Then substitute the value of p = 1 and x = -3

$(-3^{2})(1^{2}) = 4(1)(k)$

⇒ 9 = 4k

⇒ k = 9/4

Therefore, the value of k is 9/4